A stone is thrown upward from a bridge at a speed of 10 m/s. It narrowly misses the
bridge on the way back down, and hits the water at 30m/s. what is the approximate height of the bridge?
Well, I would have suggested checking the "Bridge and Stone Height" app, but apparently, that doesn't exist yet. So, let's calculate it manually.
We can solve this problem using the principles of energy conservation. At the topmost point of the stone's path, it has no kinetic energy since it momentarily stops moving before starting to fall down. Therefore, all its initial kinetic energy must have been converted into gravitational potential energy.
The kinetic energy formula is KE = (1/2) * m * v^2, and the potential energy formula is PE = m * g * h, where m is the mass, v is the velocity at the topmost point, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the bridge.
So, we have KE_initial = PE_final, which is (1/2) * m * (10 m/s)^2 = m * 9.8 m/s^2 * h.
Now, let's cancel out the masses and solve for h:
(1/2) * (10 m/s)^2 = 9.8 m/s^2 * h
25 m^2/s^2 = 9.8 m/s^2 * h
h = 25 m^2/s^2 / 9.8 m/s^2
h is approximately 2.55 meters.
Therefore, the approximate height of the bridge is approximately 2.55 meters. Just make sure to throw any clowns you encounter off the bridge to avoid any further math problems.
To find the approximate height of the bridge, you can use the concept of conservation of energy.
Initially, the stone is thrown upward from the bridge, so it has kinetic energy and potential energy.
Let's assume the kinetic energy of the stone when it is thrown upward is K1, and the potential energy is U1.
When the stone reaches its highest point, its kinetic energy becomes zero, and all its initial kinetic energy is converted into potential energy. Let's call the potential energy at its highest point U2.
When the stone comes back down and hits the water, it has potential energy and kinetic energy again. Let's call the potential energy at the water level U3, and the kinetic energy at the water level K2.
The conservation of energy equation can be written as follows:
K1 + U1 = U2 + K2 = U3
Since the stone is thrown upward and then falls back down, we can assume that the potential energy at the highest point is equal to zero.
Thus, the conservation of energy equation becomes:
K1 + U1 = K2 + U3
Given that the stone is thrown upward at a speed of 10 m/s and hits the water at a speed of 30 m/s, we can calculate the kinetic energies:
K1 = (1/2) * m * v1^2
= (1/2) * m * (10)^2
K2 = (1/2) * m * v2^2
= (1/2) * m * (30)^2
where m is the mass of the stone.
Assuming the mass of the stone cancels out when subtracting the equations, we can write:
(1/2) * (10)^2 + U1 = (1/2) * (30)^2 + U3
Simplifying the equation:
(1/2) * 100 + U1 = (1/2) * 900 + U3
50 + U1 = 450 + U3
Since the stone narrowly misses the bridge on the way back down, the height of the bridge can be approximated as the height at the highest point. Therefore, the equation becomes:
(1/2) * 100 + U1 = 450
Simplifying further:
50 + U1 = 450
U1 = 400 m
Hence, the approximate height of the bridge is 400 meters.
To approximate the height of the bridge, we can use the concept of projectile motion. When the stone is thrown upward, it goes against the force of gravity until it reaches the highest point, then it starts coming back down.
Let's assume that the stone takes t1 seconds to reach its highest point and t2 seconds to come back down to the water. We know that the initial upward velocity (v1) is 10 m/s and the downward velocity (v2) is 30 m/s.
To find the approximate height (h) of the bridge, we can use the equations of motion:
1. The height (h) can be calculated using the equation:
h = v1 * t1 - 0.5 * g * t1^2
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
2. The time taken to reach the highest point and come back down is given by:
t1 + t2 = total time of flight (t)
We can start by finding t2 using the downward velocity v2:
v2 = g * t2
Rearranging the equation:
t2 = v2 / g
Now, let's find t1 using the total time of flight (t) and t2:
t1 = t - t2
Substituting the values of v1, v2, and t into the equations, we can approximate the height of the bridge.
v^2=vo^2+2ad
30^2=10^2+2(-9.8)(d)
900=100+(-19.6)d
800=-19.6d
d=40.8